If an even function defined on R is an increasing function on [0, positive infinity), and f (2) = 1, then the x value range of F (Log1 / 2 x) > 1 is

If an even function defined on R is an increasing function on [0, positive infinity), and f (2) = 1, then the x value range of F (Log1 / 2 x) > 1 is

When x = {x | 00, the increasing function f (2) = 1, then x1;
Log1 / 2 x > 2 or Log1 / 2 x

Even function y = f (x) defined on R decreases on [0, + ∞), and f (1) 2) If = 0, then f (Log1) is satisfied The set of X with 4x < 0 is () A. (−∞，1 2)∪(2，+∞) B. (1 2，1)∪(1，2) C. (1 2，1)∪(2，+∞) D. (0，1 2)∪(2，+∞)

Because the even function y = f (x) defined on R decreases on [0, + ∞), and f (1)
2) If = 0, then f (Log1) is satisfied
4x)＜0
⇔f(|log1
4x|)＜0＝f(1
2)⇔|log1
4x|＞1
2⇔
log1
4x≥0
log1
4x＞1
2 or
log1
4x＜0
−log1
4x＞1
2 ⇒0＜x＜1
2 or x > 2
Therefore, D

The even function y = f (x) defined on R is an increasing function on the interval [0, + ∞), and a zero point of function f (x) is 1 / 2. The solution of X satisfying f (Log1 / 4 x) > 0 is obtained

If a zero point of even function f (x) is 1 / 2 and the interval [0, + ∞) is an increasing function, then f (x) > 0 means x > 1 / 2 or X1 / 2 or 2 or 0

If the function f (x) is an even function defined on R and f (x) = x 2-x when x ≥ 0, then the expression of F (x) on R is

When x > 0, - x < 0 and f (x) is an even function defined on R, then there is f (- x) = f (x) = x? - x = (- x) 2 + (- x) when x = 0, the substitution of F (x) = x? - x still holds. In conclusion, the expression of F (x) on R is x? - x x x ≥ 0, f (x) = x? + X

F (x) even function, G (x) odd function, they have the same definition domain, and f (x) + G (x) = 1 / X-1, find the expression of F (x) and G (x) As the title

f(x)+g(x)=1/x-1 .1)
f(-x)+g(-x)=1/(-x-1)
F (x) even function, G (x) odd function
So: F (x) = f (- x)
g(x)=-g(-x)
So:
f(-x)+g(-x)=1/(-x-1)
Namely:
f(x)-g(x)=1/(-x-1) .2)
1)+2):
2*f(x)=1/(x-1)-1/(x+1)=2/(x^2-1)
f(x)=1/(x^2-1)
2)
g(x)=f(x)+1/(x+1)=1/(x^2-1)+1/(x+1)=x/(x^2-1)
So: F (x) = 1 / (x ^ 2-1)
g(x)＝x/(x^2-1)

If the definition domain of F (x) and G (x) is r, f (x) is odd function, G (x) is even function, tangent f (x) + G (x) = 1 / (x ^ 2-x + 1), find the expression of F (x)

f(x)=-f(-x);g(x)=g(-x);
g(x)=1/(x^2-x+1)-f(x)
g(-x)=1/(x^2+x+1)-f(-x);
So: 1 / (x ^ 2-x + 1) - f (x) = 1 / (x ^ 2 + X + 1) - f (- x) = 1 / (x ^ 2 + X + 1) + F (x);
So: 1 / (x ^ 2-x + 1) - f (x) = 1 / (x ^ 2 + X + 1) + F (x);
After that, I can solve it by myself. When f (x) is obtained, G (x) will come out. I want to simplify it

If f (x) is even function and G (x) is odd function, they have the same definition domain, and f (x) + G (x) = 1 / X-1, find the expression of F (x), G (x)

F (x) is even function, G (x) is odd function
therefore
f(-x)＝f(x)
g(-x)＝-g(x)
f(-x)+g(-x)＝f(x)-g(x)=-1/x-1
And because
f(x)+g(x)=1/x-1
The two formulas are added together
2f(x)=-2
f(x)=-1
g(x)=1/x

Let f (x) be an odd function defined on R, G (x) be an even function defined on R, and f (x) - G (x) = 1-x2-x3, find g (x)

f(x)-g(x)=1-x^2-x^3
f(x)=g(x)+1-x^2-x^3 (1)
F (x) is an odd function defined on R and G (x) is an even function defined on R
f(-x)=-f(x);g(-x)=g(x)
According to formula (1), f (- x) = g (- x) + 1 - (- x) ^ 2 - (- x) ^ 3 = g (x) + x ^ 3-x ^ 2 + 1
-f(x)=-[g(x)+1-x^2-x^3]=-g(x)+x^2+x^3-1
That is: G (x) + x ^ 3-x ^ 2 + 1 = - G (x) + x ^ 2 + x ^ 3-1
g(x)=x^2-1
f(x)=-x^3

If the even function f (x) defined on R is from 0 to positive infinitely increasing function and f (1 / 3) = 0, then the value range of F (log x with 1 / 8 as the base) is satisfied Given that the even function f (x) defined on R increases on [0, + infinity) and f (1 / 3) = 0, then what is the value range of X satisfying f (log x with 1 / 8 base) > 0?

∵ even function f (x) is 0 to positive infinite increasing function, f (1 / 3) = 0
/ / F (x) is a minus function from negative infinity to 0, f (- 1 / 3) = 0
When x ＜ - 1 / 3 or x > 1 / 3, f (x) > 0
∵f(log(1/8) x)＞0
﹤ log (1 / 8) x ﹥ 1 / 3, or log (1 / 8) x ﹤ 1 / 3
∴0＜x＜(1/8)^(1/3)=1/2
Or X ＞ (1 / 8) ^ (- 1 / 3) = 2
The value range of X is:
{X / 0 < x < 1 / 2 or x > 2}

It is known that f (x) is an even function defined on R and an increasing function on [0, + ∞) 3) If = 0, then the inequality f (Log1 The solution set of 8x) > 0 is ⊙___ .

∵ f (x) is an even function defined on R and an increasing function on [0, + ∞)
And ∵ f (1)
3)=0，f(log1
8x)＞0
∴|log1
8x|＞1
Three
∴log1
8x＞1
3 or Log1
8x＜-1
Three
The solution is 0 < x < 1
2 or x > 2
So the answer is (0,1)
2)∪(2，+∞)．